Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $I_1=\int\limits_{e} ^{_e2}\frac{dx}{\log\,x}$ and $I_2=\int\limits_{1} ^{2}\frac{e^x}{x}dx$ , then

Integrals

Solution:

$I=\int\limits_{e}^{e^2} \frac{dx}{log\,x}$ Put log $x=z$,
$\therefore x=e^{z} $
$\therefore dx=e^{z} dz$
When $x = e, z = log e = 1$
$x=e^{2}, z=log\,e^{2}=2 \,log e=2$
$\therefore I_{1}=\int\limits_{1}^{2} \frac{e^{z}dz}{z}=\int\limits_{1}^{2} \frac{e^{x}}{z} dx=I_{2}$
$\therefore I_{1}=I_{2}$